Question Number 90520 by ajfour last updated on 24/Apr/20
Commented by ajfour last updated on 24/Apr/20
$${If}\:{eq}.\:{of}\:{ellipse}\:{is}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:, \\ $$$${find}\:{equation}\:{of}\:{shown}\:{parabola}. \\ $$
Answered by mr W last updated on 25/Apr/20
![let μ=(b/a) let eqn. of parabola be y=A(x−h)^2 +B b=A(−h)^2 +B 0=A(a−h)^2 +B ⇒(B/A)=−(a−h)^2 ⇒A=(μ/(2h−a)) ⇒B=−((μ(a−h)^2 )/(2h−a)) b^2 x^2 +a^2 [A(x−h)^2 +B]^2 −a^2 b^2 =0 b^2 x^2 +a^2 A^2 x^4 +4h^2 a^2 A^2 x^2 +a^2 A^2 h^4 −4ha^2 A^2 x^3 +2h^2 a^2 A^2 x^2 −4h^3 a^2 A^2 x+2a^2 ABx^2 −2h2a^2 ABx+2a^2 ABh^2 +a^2 B^2 −a^2 b^2 =0 A^2 x^4 −4hA^2 x^3 +(6h^2 A^2 +2AB+μ^2 )x^2 −4hA(h^2 A+B)x+(A^2 h^4 +2ABh^2 +B^2 −b^2 )=0 x=0 is a root: ⇒A^2 h^4 +2ABh^2 +B^2 −b^2 =0 ⇒A^2 x^3 −4hA^2 x^2 +(6h^2 A^2 +2AB+μ^2 )x−4hA(h^2 A+B)=0 x=a,p,p are roots: x^3 −4hx^2 +(6h^2 +2(B/A)+(μ^2 /A))x−4h(h^2 +(B/A)) =(x−a)(x−p)^2 =x^3 −(a+2p)x^2 +(2a+p)px−ap^2 (a+2p)=4h ...(i) (2a+p)p=(6h^2 +2(B/A)+(μ^2 /A)) ...(ii) ap^2 =4h(h^2 +(B/A)) ...(iii) from (i) and (iii) we get ⇒16h^2 −8ah−a^2 =0 ⇒h=(((1+(√2))a)/4)](https://www.tinkutara.com/question/Q90603.png)
$${let}\:\mu=\frac{{b}}{{a}} \\ $$$${let}\:{eqn}.\:{of}\:{parabola}\:{be} \\ $$$${y}={A}\left({x}−{h}\right)^{\mathrm{2}} +{B} \\ $$$${b}={A}\left(−{h}\right)^{\mathrm{2}} +{B} \\ $$$$\mathrm{0}={A}\left({a}−{h}\right)^{\mathrm{2}} +{B} \\ $$$$\Rightarrow\frac{{B}}{{A}}=−\left({a}−{h}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{A}=\frac{\mu}{\mathrm{2}{h}−{a}} \\ $$$$\Rightarrow{B}=−\frac{\mu\left({a}−{h}\right)^{\mathrm{2}} }{\mathrm{2}{h}−{a}} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} \left[{A}\left({x}−{h}\right)^{\mathrm{2}} +{B}\right]^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {A}^{\mathrm{2}} {x}^{\mathrm{4}} +\mathrm{4}{h}^{\mathrm{2}} {a}^{\mathrm{2}} {A}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {A}^{\mathrm{2}} {h}^{\mathrm{4}} −\mathrm{4}{ha}^{\mathrm{2}} {A}^{\mathrm{2}} {x}^{\mathrm{3}} +\mathrm{2}{h}^{\mathrm{2}} {a}^{\mathrm{2}} {A}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{4}{h}^{\mathrm{3}} {a}^{\mathrm{2}} {A}^{\mathrm{2}} {x}+\mathrm{2}{a}^{\mathrm{2}} {ABx}^{\mathrm{2}} −\mathrm{2}{h}\mathrm{2}{a}^{\mathrm{2}} {ABx}+\mathrm{2}{a}^{\mathrm{2}} {ABh}^{\mathrm{2}} +{a}^{\mathrm{2}} {B}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$${A}^{\mathrm{2}} {x}^{\mathrm{4}} −\mathrm{4}{hA}^{\mathrm{2}} {x}^{\mathrm{3}} +\left(\mathrm{6}{h}^{\mathrm{2}} {A}^{\mathrm{2}} +\mathrm{2}{AB}+\mu^{\mathrm{2}} \right){x}^{\mathrm{2}} −\mathrm{4}{hA}\left({h}^{\mathrm{2}} {A}+{B}\right){x}+\left({A}^{\mathrm{2}} {h}^{\mathrm{4}} +\mathrm{2}{ABh}^{\mathrm{2}} +{B}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:{is}\:{a}\:{root}: \\ $$$$\Rightarrow{A}^{\mathrm{2}} {h}^{\mathrm{4}} +\mathrm{2}{ABh}^{\mathrm{2}} +{B}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{A}^{\mathrm{2}} {x}^{\mathrm{3}} −\mathrm{4}{hA}^{\mathrm{2}} {x}^{\mathrm{2}} +\left(\mathrm{6}{h}^{\mathrm{2}} {A}^{\mathrm{2}} +\mathrm{2}{AB}+\mu^{\mathrm{2}} \right){x}−\mathrm{4}{hA}\left({h}^{\mathrm{2}} {A}+{B}\right)=\mathrm{0} \\ $$$${x}={a},{p},{p}\:{are}\:{roots}: \\ $$$${x}^{\mathrm{3}} −\mathrm{4}{hx}^{\mathrm{2}} +\left(\mathrm{6}{h}^{\mathrm{2}} +\mathrm{2}\frac{{B}}{{A}}+\frac{\mu^{\mathrm{2}} }{{A}}\right){x}−\mathrm{4}{h}\left({h}^{\mathrm{2}} +\frac{{B}}{{A}}\right) \\ $$$$=\left({x}−{a}\right)\left({x}−{p}\right)^{\mathrm{2}} \\ $$$$={x}^{\mathrm{3}} −\left({a}+\mathrm{2}{p}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{a}+{p}\right){px}−{ap}^{\mathrm{2}} \\ $$$$\left({a}+\mathrm{2}{p}\right)=\mathrm{4}{h}\:\:\:\:…\left({i}\right) \\ $$$$\left(\mathrm{2}{a}+{p}\right){p}=\left(\mathrm{6}{h}^{\mathrm{2}} +\mathrm{2}\frac{{B}}{{A}}+\frac{\mu^{\mathrm{2}} }{{A}}\right)\:\:\:…\left({ii}\right) \\ $$$${ap}^{\mathrm{2}} =\mathrm{4}{h}\left({h}^{\mathrm{2}} +\frac{{B}}{{A}}\right)\:\:\:…\left({iii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({iii}\right)\:{we}\:{get} \\ $$$$\Rightarrow\mathrm{16}{h}^{\mathrm{2}} −\mathrm{8}{ah}−{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{h}=\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){a}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 25/Apr/20
Commented by mr W last updated on 25/Apr/20
Commented by mr W last updated on 25/Apr/20
Commented by ajfour last updated on 25/Apr/20
$$\mathcal{B}{eautiful},\:{Sir}.\:\mathcal{T}{hanks}\:{a}\:{lot}! \\ $$
Answered by ajfour last updated on 25/Apr/20
![let eq. of parabola be y=c(x−a)(x−(b/(ac))) ellipse: b^2 x^2 +a^2 y^2 =a^2 b^2 for their common points, which are x=0, x=a, x=p, x=p b^2 x^2 +a^2 c^2 (x−a)^2 (x−(b/(ac)))^2 =a^2 b^2 ⇒ a^2 c^2 x^4 −2a^2 c^2 (a+(b/(ac)))x^3 +a^2 c^2 [((2b)/c)+(b^2 /(a^2 c^2 ))+(a+(b/(ac)))^2 ]x^2 −2a^2 bc(a+(b/(ac)))x = 0 Now a+2p=2(a+(b/(ac))) and ap^2 =((2b)/c)(a+(b/(ac))) ⇒ a(a+((2b)/(ac)))^2 =((8b)/c)(a+(b/(ac))) ⇒ (a^2 c+2b)^2 =8b(a^2 c+b) or a^4 c^2 −4a^2 bc−4b^2 =0 ⇒ c=((2b)/a^2 )(1+(√2)) , hence eq. of parabola is y=((2b)/a^2 )(1+(√2))(x−a)[x−(b/a)∙(a^2 /(2b(1+(√2))))] or y=2(1+(√2))((b/a^2 ))(x−a){x−((a((√2)−1))/2)}](https://www.tinkutara.com/question/Q90672.png)
$${let}\:{eq}.\:{of}\:{parabola}\:{be} \\ $$$$\:{y}={c}\left({x}−{a}\right)\left({x}−\frac{{b}}{{ac}}\right) \\ $$$${ellipse}:\:\:{b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${for}\:{their}\:{common}\:{points}, \\ $$$${which}\:{are}\:{x}=\mathrm{0},\:{x}={a},\:{x}={p},\:{x}={p} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \left({x}−{a}\right)^{\mathrm{2}} \left({x}−\frac{{b}}{{ac}}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} {c}^{\mathrm{2}} {x}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} {c}^{\mathrm{2}} \left({a}+\frac{{b}}{{ac}}\right){x}^{\mathrm{3}} \\ $$$$\:\:\:\:+{a}^{\mathrm{2}} {c}^{\mathrm{2}} \left[\frac{\mathrm{2}{b}}{{c}}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} {c}^{\mathrm{2}} }+\left({a}+\frac{{b}}{{ac}}\right)^{\mathrm{2}} \right]{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:−\mathrm{2}{a}^{\mathrm{2}} {bc}\left({a}+\frac{{b}}{{ac}}\right){x}\:=\:\mathrm{0} \\ $$$${Now}\:\:\:\:{a}+\mathrm{2}{p}=\mathrm{2}\left({a}+\frac{{b}}{{ac}}\right) \\ $$$$\:\:{and}\:\:\:\:\:\:{ap}^{\mathrm{2}} =\frac{\mathrm{2}{b}}{{c}}\left({a}+\frac{{b}}{{ac}}\right) \\ $$$$\Rightarrow\:\:{a}\left({a}+\frac{\mathrm{2}{b}}{{ac}}\right)^{\mathrm{2}} =\frac{\mathrm{8}{b}}{{c}}\left({a}+\frac{{b}}{{ac}}\right) \\ $$$$\Rightarrow\:\:\left({a}^{\mathrm{2}} {c}+\mathrm{2}{b}\right)^{\mathrm{2}} =\mathrm{8}{b}\left({a}^{\mathrm{2}} {c}+{b}\right) \\ $$$${or}\:\:\:\:\:\:{a}^{\mathrm{4}} {c}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {bc}−\mathrm{4}{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\:{c}=\frac{\mathrm{2}{b}}{{a}^{\mathrm{2}} }\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:\:\:,\:\:{hence} \\ $$$$\:\:\:{eq}.\:{of}\:{parabola}\:{is} \\ $$$$\:\:{y}=\frac{\mathrm{2}{b}}{{a}^{\mathrm{2}} }\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left({x}−{a}\right)\left[{x}−\frac{{b}}{{a}}\centerdot\frac{{a}^{\mathrm{2}} }{\mathrm{2}{b}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}\right] \\ $$$${or} \\ $$$${y}=\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\frac{{b}}{{a}^{\mathrm{2}} }\right)\left({x}−{a}\right)\left\{{x}−\frac{{a}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}}\right\}\: \\ $$
Commented by ajfour last updated on 25/Apr/20
